Different A.P.'s are constructed with the first term 100, the last term 199, And integral common differences. The sum of the common differences of all such, A.P's having at least 3 terms and at most 33 terms is.

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answer is 53.

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Detailed Solution

$1^{st} term = 100 = a$

Last term = 199 = l

$\begin{array}{l} a_{n} = ℓ = a + (n - 1) d \\ d_{i} = \frac{ℓ - a}{n - 1} \end{array}$

n $\to$ number of terms

$\begin{array}{l} n = 3, d_{1} = \frac{199 - 100}{2} \\ = \frac{99}{2} \notin I \\ n = 4, d_{2} = \frac{99}{3} = 33 \in I \\ n = 10, d_{3} = \frac{99}{9} = 11 \in I \\ n = 12, d_{4} = \frac{99}{11} = 9 \in I \\ ∴ \sum d_{i} = 33 + 11 + 9 = 53 \end{array}$

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